Let S be the set of all complex numbers z satisfying z - 2 + i ≥ 5 . If the complex number z 0 is such that 1 z 0 - 1 is the maximum of the set 1 z - 1 : z ∈ S , then the principal argument of 4 - z 0 - z ¯ 0 z 0 - z ¯ 0 + 2 i is
D. - π 2
The region represented by z - 2 + i ≥ 5 will be region outside and on circle with centre 2 , - 1 and radius 5 . z - 2 + i ≥ 5 Let z = x + i y ⇒ x + i y - 2 + i ≥ 5 ⇒ x - 2 + i y + 1 ≥ 5 ⇒ x - 2 2 + y + 1 2 ≥ 5 ⇒ x - 2 2 + y + 1 2 ≥ 5 Now, for 1 z 0 - 1 to be maximum, z 0 - 1 must be minimum. We need to find the B z 0 which is in given region and nearest to point A 1 , 0 , hence nearest point from A 1 , 0 will be on the line joining A and C . Method 1 Let z 0 = x + i y , then x 2 and y > 0 (from diagram) Consider w = 4 - z 0 - z ¯ 0 z 0 - z ¯ 0 + 2 i = 4 - x + i y - x - i y x + i y - x - i y + 2 i = 4 - 2 x i y + 2 = 1 i 4 - x y + 2 w = - i 4 - x y + 2 w = i K , where K = 4 - x y + 2 As x 2 and y > 0 ⇒ K > 0 w = - i K , will lie on negative imaginary axis A r g w = - π 2 Alternate: As B z 0 lies on line A C , Equation of A C : y + 1 = 0 - - 1 1 - 2 x - 2 ⇒ y + 1 = 1 - 1 x - 2 ⇒ x + y = 2 . . . . i Let z 0 = x + i y Consider w = 4 - z 0 - z ¯ 0 z 0 - z ¯ 0 + 2 i w = 4 - x i y + 2 From equation i w = 4 - x i 2 - x + 2 w = 1 i w = - i A r g w = - π 2
Related: Mathematics — Complex Number · All PYQ Banks